SAJC H1 CHEM P1 Worked solutions

1  ST ANDREW’S JUNIOR COLLEGE JC2 Preliminary Examination (worked solutions) Chemistry Higher 1 8872/01 Paper 1 Multiple Choice 19 September 2016 50 minutes Additional Materials: Multiple Choice Answer Sheet, Data Booklet READ THESE INSTRUCTIONS FIRST Write in soft pencil. Do not use staples, paper clips, highlighters, glue or correction fluid. There are 30 questions on this paper. Answer a ll questions. For each question there are four possible ans wers A, B, C and D. Choo se the one you consider correct and record your choice in soft pencil on the separate Answer Sheet. Each correct answer will score one mark. A mark will not be deducted for a wrong answer. Any rough working should be done in this booklet. This document consists of 13 printed pages including this page.

2  Section A 1 Percentage by mass of C in ethanol = (12 x 2) / [(12 x 2) + 6 + 16] = 0.5217 Thus, Mass of C in 11.5 g ethanol = 0.5217 x 11.5 = 6g Therefore, No. of moles of C in 11.5 g ethanol = 6/12 = 0.5 No. of carbon atoms in 11.5 g ethanol = 0.5 x L = 0.5L or L/2 Answer: B 2   C H Cl % 24.2 4.1 71.2 Divide by Ar 24.2 / 12 = 2 4.1 / 1 = 4.1 71.2 / 35.5 = 2 Divide by smallest number 2 / 2 =1 4.1 / 2 = 2 2 / 2 =1 Ratio 1 2 1 Therefore, empirical formula of CFC-150a is CH2Cl Among the options, only D has this empirical formula Answer: D 3 A: No change in oxidation number (O.N) of S, remaining at +6 B: O.N of S changed from +6 to +4 C: O.N of S changed from +6 to 0 D: O.N of S changed from +6 to -2, thus the biggest difference/change Answer: D

3  4 Since P and Q are in the same period, - P must come after Q to have more protons than Q - To be after Q and have a lower 1st IE, P must be either Gp III (and Q Gp II) or Gp V (and Q Gp VI) as these are where the anomalies of the increasing 1 st IE trend across the period. - If Q is Gp II and P is Gp III, P has one more unpaired electron than Q as the valence electron configuration of P : Q would be ns2 : ns2 np1; If Q is Gp V and P is Gp VI, P has one less unpaired electron than Q as the valence electron configuration of P : Q would be ns2 np4: ns2 np3 Thus, P must be in Gp VI while Q is in Gp V Answer: C 5 A: Ca(NO3)2: There is ionic bonds between Ca 2+ and NO3- as well as covalent bonds within the NO3- anion. B: MgS: There is only ionic bonds between Mg2+ and S2-. C: Only contains metallic bonds given that it’s a metal D: SO2: Only contains covalent bonds given that it’s a giant covalent compound Answer: A 6 Full structure of compound R is   The bond angles present are: - 109.5° around C that have 4 sigma bonds - 180° around the C of CN - 120° around the C of C=C - 105° around the O of O-H Answer: B

4  7 A: False. Its not true as the order should be 4>3>1>2 as 2 is branched while 1 is straight chain so 1 should have more extensive id-id due to larger surface area when compared with 2. B: False. As mentioned above, 2 is branched while 1 is straight chain so 1 should have more extensive id-id due to larger surface area when compared with 2, thus their melting point is not the same. C: False. Predominant force for 3 is pd-pd but 4 is hydrogen bonding so different D: True. Both 1 and 2 are non-polar so predominant interaction is id-id. Answer: D 8 The ∆H of neutralisation between a strong acid and a strong base, as shown by the first example in the table above is -57 kJ. Thus, for the values to be less exothermic than -57, it would mean that J must have been a weak acid (given that KOH is a strong base) and L is a weak base (given that nitric acid is a strong acid). For the last one, the value is twice of that of -57. Hence it must be that M is a dibasic acid reacting with a diacidic base like Ca(OH) 2 Thus, J has to be propanoic acid, L is ethylamine and M is sulfuric acid Ans: C 9 If enthalpy change of formation is given, ∆H of any reaction = Σ∆Hf (products) - Σ∆Hf (reactants) Thus, +70 = [∆Hf (Cl2)+ (2 x ∆Hf (O2))] – (2 x∆Hf (ClO2)) = 0 + 0 - (2 x∆Hf (ClO2)) [since Cl2 and O2 are both elements, thus ∆Hf = 0] Therefore, 2 x∆Hf (ClO2) = - 70 ∆Hf (ClO2) = -35 kJ mol-1 Ans: B

5  10 When concentration of E was halved, the rate would be halved, given that its first order with respect to E. Hence, to get the overall rate to be doubled, F needs to increase by a factor of 2 so that the rate will be increase by 4 times (as its second order with respect to F) Ans: C 11 For A, graph is  or   Thus its not a straight line For B, graph is or   Thus its a straight line

6  For C, graph is   Thus its a straight line For D, graph is   Thus its a straight line Ans: A 12 Homogenous gaseous system implies all reactants and products are gases. Thus, when pressure is increase and equilibrium shift to the right, it means that there are less gaseous molecules on the right so that pressure can be reduced. Since Kc is [products] x / [reactants] y, x is less than y for this system resulting in units of such an equilibrium to be (moldm-3) -z resulting in units being mol-zdm3z. Ans: A

7  13 Both acids are strong acids that dissociate completely. Therefore, Total no. of moles of H+ ions = [(50/1000) x 0.02] + [(150/1000) x 0.03] = 0.0055 Final concentration of H+ ions = 0.0055 / (200/1000) = 0.0275 Therefore, pH = -log (0.0275) = 1.56 Ans: C   14 pH of such a solution = -log (1.00 x 10-6) = 6 Thus, according to the table given, solution will appear blue in bromocresol-green as the pH of the solution is above its colour change pH range and yellow for phenol-red as the pH of the solution is below its colour change pH range. Ans: B 15 For A, though more H+ ion is produced at 50°C as equilibrium shift to the right. However, the same amount of OH- ions is also produced making water still neutral and not more acidic. Hence false. For B, at 50°C, equilibrium shift to the right and hence the Kw value will be larger than 1 x 10-14. This explains why option D is false as well. Thus, the square root of a larger value to obtain the concentration of OH- will be larger not smaller than 1 x 10-7. Hence false. For C, following the explanation for B, the H+ concentration will be larger than 1 x 10-7 as well, resulting in a pH (which is the –log of a value larger than 1 x 10-7) smaller than 7. Hence true. Ans: C

8  16 For A, its decreasing as atomic radius decrease across the period. For B, its decreasing as well as electrical conductivity decreases across the period from Na to P. For C, its also decreasing as ionic radius of cations decreases as long as they are in the same period. For D, although melting point decrease from Si to P, but S has a higher melting point than P as the number of electrons for S is more in the form of S 8 as compared to P in the form of P4. This results in stronger id-id for S than for P. Hence its not decreasing. Ans: D 17 For method 1, after heating both in oxygen, Al2O3 and SiO2 will be obtained. Only Al2O3 will react with acid as it is amphoteric but SiO2 will have no reaction with water. Hence method 1 is able to help identify them. For method 2, after heating both in chlorine, AlCl3 and SiCl4 will be obtained. Due to the aqueous nature of the base, AlCl3 will dissolve aqueous ions but SiCl4 will give solid SiO2.2H2O, which does not dissolve in a base. Hence method 2 can also help to identify them. Ans: C 18      : sp2 hybrdised : sp3 hybrdised Thus there are 5 sp2 hybridised carbon atoms and 5 sp3 hydridised carbon atoms ANs: C

9  19 The following shows the labelled diagram of 1,3-dimethylcyclohexane where chlorine can substitute to result in mono-substituted chlorinated isomers. Carbons that show the same product are labelled as the same carbon as well.   Ans: C 20   End product: sodium propanoate Ans: D 21 The precipitate after adding AgNO3 is AgCl. No. of moles of AgCl = 28.7 / (108+35